temp-> curlen=len;/子串长度 for (int i=0,j=pos; i< len; 1++,j++) temp->chi=chj];传送串数组 temp-> chen]=“0,;子串结束 return x temp; example:st=“ university”,pos=3,len=4 use case: subSt=st (3, 4 substring extraction: subst=“vers”temp->curLen = len; //子串长度 for ( int i = 0, j = pos; i < len; i++, j++ ) temp->ch[i] = ch[j]; //传送串数组 temp->ch[len] = ‘\0’; //子串结束 } return * temp; } example：st = “university”, pos = 3, len = 4 use case: subSt = st (3, 4) substring extraction: subSt = “vers